#2087

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Cost Homecoming of a Robot in a Grid

Find the minimum cost for a robot to return home in a grid with row and column movement costs.

Problem Statement

You are given an m x n grid where (0, 0) is the top-left cell and (m - 1, n - 1) is the bottom-right cell. A robot starts at a given position startPos = [startrow, startcol] and needs to reach homePos = [homerow, homecol]. The robot can move up, down, left, or right, but cannot move outside the grid. Each move has a cost associated with it, defined by the arrays rowCosts and colCosts, where rowCosts[r] specifies the cost of moving along row r and colCosts[c] specifies the cost of moving along column c.

Your task is to compute the minimum total cost for the robot to reach its home position from the starting position.

Examples

Example 1

Input: startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]

Output: 18

One optimal path is that: Starting from (1, 0) -> It goes down to (2, 0). This move costs rowCosts[2] = 3. -> It goes right to (2, 1). This move costs colCosts[1] = 2. -> It goes right to (2, 2). This move costs colCosts[2] = 6. -> It goes right to (2, 3). This move costs colCosts[3] = 7. The total cost is 3 + 2 + 6 + 7 = 18

Example 2

Input: startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]

Output: 0

The robot is already at its home. Since no moves occur, the total cost is 0.

Constraints

m == rowCosts.length
n == colCosts.length
1 <= m, n <= 105
0 <= rowCosts[r], colCosts[c] <= 104
startPos.length == 2
homePos.length == 2
0 <= startrow, homerow < m
0 <= startcol, homecol < n

Solution Approach

Greedy Path Calculation

The key observation is that the robot will traverse all the rows between the start row and the home row, and all the columns between the start column and the home column. Therefore, the robot must incur the sum of the row and column movement costs. This greedy approach ensures that each step is optimized to minimize the cost of moving in both dimensions, row by row and column by column.

Efficient Cost Calculation

Once the robot starts its journey from startPos, we simply sum the costs for traversing the required rows and columns. This can be done by iterating over the relevant rows and columns, accumulating the costs for each movement. This greedy approach is direct and avoids unnecessary complexity by focusing on the minimal required movements.

Edge Case Handling

If the robot is already at home (startPos == homePos), the total cost is zero, as no moves are needed. This edge case is important to handle early in the solution to avoid unnecessary calculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is determined by the number of rows and columns the robot needs to traverse. In the worst case, the robot will traverse all the rows between startRow and homeRow, and all the columns between startCol and homeCol, leading to a time complexity of O(m + n). The space complexity is O(1), as we only need a few variables to accumulate the total cost.

What Interviewers Usually Probe

Understanding of greedy algorithms for pathfinding in grid-based problems.
Ability to identify edge cases and handle them efficiently.
Focus on optimizing space and time complexity in a problem involving traversal.

Common Pitfalls or Variants

Common pitfalls

Failing to handle the case where startPos equals homePos, leading to unnecessary calculations.
Not recognizing the grid structure and movement constraints, which may lead to overcomplicating the solution.
Misunderstanding the problem as requiring a complex pathfinding algorithm when it's a simple greedy approach.

Follow-up variants

Consider a grid with additional constraints like obstacles or variable costs for each cell.
Extend the problem to a 3D grid with movement in three dimensions.
What if the robot could only move diagonally or had some restricted movement pattern?

FAQ

What is the main algorithmic approach to solving Minimum Cost Homecoming of a Robot in a Grid?

The problem is solved using a greedy approach, where the robot's movement cost is calculated by directly summing the row and column costs for the necessary movements.

How do you handle edge cases where the robot is already at home?

If the robot is already at home, no movement is needed, so the total cost is zero, which is handled as a special case.

What is the time complexity of the solution for this problem?

The time complexity is O(m + n), where m and n are the dimensions of the grid, because the robot needs to traverse all rows and columns between the start and home positions.

Can this problem be solved with a dynamic programming approach?

Although dynamic programming could be applied, this problem is better suited to a greedy approach because of the fixed movement costs and the simplicity of the grid traversal.

What would happen if obstacles were added to the grid?

If obstacles were added, the problem would require a more complex pathfinding algorithm, such as Dijkstra's algorithm or A* search, to account for blocked paths.

terminal

Solution

Solution 1: Greedy

Let's assume the robot's initial position is $(x_0, y_0)$ and the home position is $(x_1, y_1)$.

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class Solution:
    def minCost(
        self,
        startPos: List[int],
        homePos: List[int],
        rowCosts: List[int],
        colCosts: List[int],
    ) -> int:
        x0, y0 = startPos
        x1, y1 = homePos
        dx = sum(rowCosts[x0 + 1 : x1 + 1]) if x0 < x1 else sum(rowCosts[x1:x0])
        dy = sum(colCosts[y0 + 1 : y1 + 1]) if y0 < y1 else sum(colCosts[y1:y0])
        return dx + dy

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